Problem: Find $\lim_{x\to 0}\dfrac{\sin(x)\cos(x)}{x+\sin(2x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $1$ (Choice C) C $\dfrac{1}{3}$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=0$ into $\dfrac{\sin(x)\cos(x)}{x+\sin(2x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{\sin(x)\cos(x)}{x+\sin(2x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\sin(x)\cos(x)]}{\dfrac{d}{dx}[x+\sin(2x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\cos^2(x)-\sin^2(x)}{1+2\cos(2x)} \\\\ &=\dfrac{\cos^2(0)-\sin^2(0)}{1+2\cos(2\cdot0)} \gray{\text{Substitution}} \\\\ &=\dfrac{1}{3} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\sin(x)\cos(x)]}{\dfrac{d}{dx}[x+\sin(2x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{\sin(x)\cos(x)}{x+\sin(2x)}=\dfrac{1}{3}$.